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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Chapter 6. Dynamics I: Motion Along a Line Chapter Goal: To learn how to solve problems about motion in a straight line.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Student Learning Objectives – Ch. 6 To draw and make effective use of free-body diagrams. To recognize and solve simple equilibrium problems. To distinguish mass, weight, and apparent weight. To learn and use simple models of friction. To apply the full strategy for force and motion problems to problems in single-particle dynamics.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Equilibrium An object on which the net force is zero is said to be in equilibrium. Static equilibrium: object is at rest. Dynamic equilibrium: moving along a straight line with constant velocity. Both are identical from a Newtonian perspective because the net force and the acceleration are zero.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Problem-Solving Strategy: Equilibrium Problems

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Problem-Solving Strategy: Equilibrium Problems

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example Equilibrium Problem (#31) A 500 kg piano is being lowered into position by a crane while 2 people steady it with ropes pulling to the side. Bob’s rope pulls left, 15 0 below horizontal, with 500 N of tension. Ellen’s rope pulls right, 25 0 below horizontal. a.What tension must Ellen maintain in her rope to keep the piano descending at a steady speed? b.What is the tension in the main cable?

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example Problem (#31) – Freebody Diagram T3T3 T 1 is Bob’s side, T 2 is Ellen’s side, T 3 is the main cable. Same system for the angles

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Problem #31 - Apply Newton’s 1 st Law ΣF y = 0, ΣF x = 0 T3T3

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Problem #31 – Solve and assess T 2 = 533 N, T 3 = 5.25 x 10 3 N. The cable must supports the weight of the piano (4900 N) plus the added downward components of the tension in the supporting ropes.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. When your car rapidly accelerates forward, you are pressed into the seat. When your car suddenly comes to a halt, you lunge forward. These two phenomena are an example of: A.Newton’s 1 st Law; the sum of the forces upon you is zero. B.Newton’s 2 nd Law; you experience a net force and you accelerate C.Newton’s 3 rd Law; you experience a force equal and opposite to that of the car

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. When your car rapidly accelerates forward, you are pressed into the seat. When your car suddenly comes to a halt, you lunge forward. These two phenomena are an example of: A.Newton’s 1 st Law; the sum of the forces upon you is zero. B.Newton’s 2 nd Law; you experience a net force and you accelerate C.Newton’s 3 rd Law; you experience a force equal and opposite to that of the car

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Problem-Solving Strategy for Newton’s 2 nd Law Problems 1.Use the problem-solving strategy outlined for Newton’s 1 st Law problems to draw the free body diagram and determine known quantities. 2.Use Newton’s Law in component form to find the values for any individual forces and/or the acceleration. 3. If necessary, the object’s trajectory (time, velocity, position, acceleration) can be determined by using the equations of kinematics. 4.Reverse # 2 and 3 if necessary.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. For which of the following is ma y = F 1 -F 2 cosθ - F 3 sin θ ABAB CDCD

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. For which of the following is ma y = F 1 -F 2 cosθ - F 3 sin θ ABAB CDCD

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example Dynamics Problem A 75-kg snowboarder starts down a 50-m high, 10 0 slope on a frictionless board. What is his speed at the bottom?

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example Dynamics Problem – Visualize Freebody Diagram A 75-kg snowboarder starts down a 50-m high, 10 0 slope on a frictionless board. What is his speed at the bottom? Find v 1 a n FGFG

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example Dynamics Problem – Newton’s 2 nd Law in component form to solve for acceleration A 75-kg skier starts down a 50- m high, 10 0 slope on a frictionless board. What is his speed at the bottom? n FGFG ΣF y = ma y = 0 ΣF x = ma x a = 1.7 m/s 2 Supports earlier statement that a = g sinθ

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example Dynamics Problem – Use kinematics to find speed. Is time important? A 75-kg snowboarder starts down a 50-m high, 10 0 slope on a frictionless board. What is his speed at the bottom? Note (the slope is 50 m high, not long!) v 1 = 31.3 m/s. That’s about 60 mph! Find v 1 a = 1.7 m/s 2, from previous

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Use the graph to answer the question #1 The figure shows a force acting on a 2.0-kg object moving along the x-axis. The object is at rest at the origin at t=0. What is the acceleration of the object at t = 2s? A.4 m/s 2 B.2 m/s 2 C.8 m/s 2 D.0 m/s 2

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Use the graph to answer the question #1 The figure shows a force acting on a 2.0-kg object moving along the x-axis. The object is at rest at the origin at t=0. What is the acceleration of the object at t = 2s? A.4 m/s 2 B.2 m/s 2 C.8 m/s 2 D.0 m/s 2

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Use the graph to answer the question #2 The figure shows a force acting on a 2.0-kg object moving along the x- axis. The object is at rest at the origin at t=0. What is the velocity of the object at t = 6s? A. v = 4m/s B. v = -1 m/s C.v = 0 m/s D.v = 2 m/s

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Use the graph to answer the question #2 The figure shows a force acting on a 2.0-kg object moving along the x- axis. The object is at rest at the origin at t=0. What is the velocity of the object at t = 6s? A. v = 4m/s B. v = -1 m/s C.v = 0 m/s D.v = 2 m/s

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Mass and Gravity Mass is a scalar quantity that describes the amount of matter in an object. Mass is an intrinsic property of an object. The force of gravity is an attractive, long-range “inverse square” force between any two objects.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The figure shows the moon (m 1 ) and the earth (m 2 ). The earth is approximately 80 times as massive as the moon. The red arrow shown is the force that the earth exerts on the moon (F 2on1 ). The moon also exerts a force on the earth, F 1on2, shown in blue (not to scale!). The magnitude of this force is: a. about 80 smaller than F 2on1 b. somewhat smaller than F 2on1 c. Equal to F 2on1 d. Not related to F 2on1. The Earth and the Moon moon earth ?

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The figure shows the moon (m 1 ) and the earth (m 2 ). The earth is approximately 80 times as massive as the moon. The red arrow shown is the force that the earth exerts on the moon (F 2on1 ). The moon also exerts a force on the earth, F 1on2, shown in blue (not to scale!). The magnitude of this force is: c. Equal to F 2on1 The Earth and the Moon moon earth

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Consider an object of mass m, on or near the surface of a planet. We can write the gravitational force even more simply as: where the quantity g is defined to be M, R represent the mass and radius of the planet. The weight force is not an intrinsic property of an object and does not have a unique value. The direction of the gravity vector defines true vertical. gravitational (weight )force

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Apparent Weight Apparent weight w, is a contact force (e.g. T, n, or F sp ), which can be thought of as “what the scale says”, although there is not always a scale. If object and scale are in vertical static or dynamic equilibrium w = F G = mg. If object and scale accelerate vertically, w ≠ mg. It must be calculated using Newton’s 2 nd Law. The use of w instead of n or T is optional, as long as you know which force is the apparent weight.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. An elevator suspended by a cable is moving upward and slowing to a stop. As it does, your apparent weight is: A. less than your true weight, which is mg. B. equal to your true weight, which is mg. C. more than your true weight, which is mg. D. zero.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. An elevator suspended by a cable is moving upward and slowing to a stop. As it does, your apparent weight is: A. less than your true weight, which is mg. B. equal to your true weight, which is mg. C. more than your true weight, which is mg. D. zero. n

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Workbook Problem # 18

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Workbook Problem # 18 - ans Explanation: S is the normal force which is the apparent weight. From N’s 2 nd Law: S= ma + |mg| speed and direction are not relevant.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Apparent Weight Problem A 50-kg woman gets in a 1000-kg elevator at rest. The elevator has a scale in it (I hate when that happens). As the elevator begins to move, the scale reads 600 N for the first 3 seconds. a.Can you tell which direction she moved? If so, what is it? b.How far has the elevator moved in those 3 s?

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Apparent Weight Problem (a) Calculate the true weight (mg) and draw a free-body diagram that correctly shows relative vector lengths In this case the scale reads heavy (I hate when that happens) so net force and acceleration are in the positive direction. Elevator must be going up and speeding up or going down and slowing down. Only one is possible due to initial velocity constraint. known: v 0 = 0 m/s m = 50 kg n = 600 N g = 9.8 m/s 2

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Apparent Weight Problem (b) How far has the elevator moved in those 3 s? The pictorial representation shows not enough information to solve problem, so we go to Newton’s 2 nd Law analysis y 0 = t 0 = v 0 = 0 a0a0 y 1, t 1 = 3s v 1 0 m

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Apparent Weight Problem (b) ΣF y = ma y using known values: n – mg = ma a = 2.2 m/s 2 a known: m = 50 kg n = 600 N g = 9.8 m/s 2

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Apparent Weight Problem (b) How far has the elevator moved in those 3 s? Time is important so use position equation with v 0 = 0m/s: ∆ y = ½ a ∆t 1 2 ∆ y = 9.9 m y 0 = t 0 = v 0 = 0 a 0 = 2.2 m/s 2, determined from N’s 2 nd Law y 1, t 1 = 3s v 1 0 m

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Friction

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Kinetic Friction Experiments show that the kinetic friction force is nearly constant and proportional to the magnitude of the normal force. where the proportionality constant μ k is called the coefficient of kinetic friction (table in text).

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Static Friction The box is in static equilibrium, so the static friction must exactly balance the pushing force:

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Static friction An object remains at rest as long as f s < f s max The object slips when f s = f s max A static friction force f s > f s max is not physically possible. f s max is always >f k for the same surfaces where the proportionality constant μ s is called the coefficient of static friction.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Comparison of static and kinetic friction

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Rolling Friction Rolling friction acts much like kinetic friction, but values for u r are much less than those for u k. Rolling friction is a resistive force. It is not the same as the static friction that provides the propulsion force that move the wheel forward.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A model of friction “ motion” indicates motion relative to the two surfaces the max value static friction, f s max occurs at the very instant the object begins to move (which often means 1 ns before, for problem-solving purposes.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Rank in order, from largest to smallest, the magnitude of the friction forces in these five different situations. The box and the floor are made of the same materials in all situations. The push force is not necessarily the same.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Rank order, from largest to smallest, the size of the friction forces in these five different situations. The box and the floor are made of the same materials in all situations. f b > f c = f d = f e > f a.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example kinetic friction problem A 75-kg snowboarder starts down a 50-m high, 10 0 slope with μ k = 0.06. What is his speed at the bottom? This is the same problem as before only the slope is no longer frictionless. Before, the velocity was 31.3 m/s. How does friction change that?

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example kinetic friction problem A 75-kg snowboarder starts down a 50-m high, 10 0 slope μ s = 0.12 and μ k = 0.06 What is his speed at the bottom? n FGFG fkfk Find v 1

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example kinetic friction problem A 75-kg snowboarder starts down a 50-m high, 10 0 slope on a frictionless board. What is his speed at the bottom? v 1 = 25.4 m/s. Compare with “frictionless” problem. Friction acts to slow him down, although not by much. Find v 1 a, from previous

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example “max” problem A truck is hauling a crate when it starts up a 10.0˚ hill. The coefficients of friction are μ s = 0.35, and μ k = 0.15, respectively. What is the maximum acceleration the truck can have as he goes up the hill, without the crate slipping backward?

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example static friction problem n FgFg fsfs θ knownfind θ = 10˚a max u s =.35 u k =.15 When does a max occur? Find n using Newton’s 2 nd law in the y direction. Find a max using Newton’s 2 nd law in the x direction.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example static friction problem n FgFg fsfs θ knownfind θ = 10˚a max u s =.35 u k =.15 a max =1.68 m/s/s

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Keep the picture up A person is trying to judge whether a picture of mass 1.10 kg is properly positioned by pressing it against a wall. The pressing force is perpendicular to the wall. The coefficient of static friction between picture and wall is 0.660. What is the minimum amount of pressing force required? Draw a freebody diagram. In which direction is the normal force in this problem? Does it have anything to do with the weight of the picture?

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Keep the picture up Freebody diagram F push FgFg fsfs n fbd - picture Knowns m = 1.10 kg μ s = 0.660 Find F push Forces which are usually x are y in this problem and vice versa (with the exception of gravity). Newton’s Laws still work.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Keep the picture up Freebody diagram F push FgFg fsfs n fbd - picture Knowns m = 1.10 kg μ s = 0.660 Find F push Newton’s Law in the x direction tells us that n = F push but nothing else about the value of either. Moving right along to the y- direction: ΣF y = ma y = 0 = f s – F G or f s = mg. No matter how hard you press, the picture will not levitate up. Fact. However, if you don’t push hard enough….

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Keep the picture up - answer F push FGFG fsfs n fbd - picture Knowns m = 1.10 kg μ s = 0.660 Find F push The minimum value of n must be the value that allows f smax to be equal to the weight of the picture: ΣF y = 0 = f smax – F G or μ s |n| = mg n = F push = 16.3 N

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Using Newton’s 2 nd Law: Workbook exercises Answers: 7-12

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